Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 4 - Motion in Two and Three Dimensions - Problems: 16b

Answer

t = 0.75s

Work Step by Step

In the previous section we found that a = $(6-8t)i $ Therefore, to find when the acceleration is 0, we plug in 0 for a and solve for the t at which this occurs. The j and k components are always 0 so we need only worry about the i component. The magnitude of this vector is given by: $magnitude = \sqrt(i^2 + j^2 +k^2) $ where i, j, and k represent the i, j, and k components respectively Therefore, we have $ magnitude = \sqrt((6-8t)^2) $ which is just 6-8t Therefore we can solve for when a = 0: $ 0 = 6-8t $ so then t = 0.75s
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